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Shear stresses in pairs act on an element at or near the wide surface, as shown in Fig. 4-4a, As shown in strength of materials , this state of stress corresponds to equal tension and compression stresses on the faces of an element at 45°to the direction of shear . These inclined tension stresses are of the same kind as those cause by transverse shear .However, in the case of torsion, since the torsional shear stresses are of opposite signs in the two halves of the member(Fig.4-4b)the corresponding diagonal tension stresses in the two halves are at right angles to each other(Fig.4-4a). When the diagonal tension stresses exceed the tension resistance of the concrete, a crack forms at some accidentally weaker location and spreads immediately across the beam, as shown in Fig.4-5,Observation shows that the tension crack(on the near face on Fig.4-5b) forms at practically 45°,that is, perpendicular to the diagonal tension stresses. The cracks on the two narrow faces, where diagonal tension stresses are smaller, are of more indefinite inclination, as shown, and the fracture line on the far face connects the cracks at the short faces. This completes the formation of an entire fracture surface across the beam which fails the member.